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3 years ago

Pls help me and show working

Mathematics

1 answer:

Juliette [100K]3 years ago

6 0

Answer:

4x + 10 > 3x + 14

x > 4

Step-by-step explanation:

Use the given values for length and width in the formula for perimeter of a rectangle. You will be able to find the perimeter of rectangles A and B.

A = 2(l + w)

A = 2(x + 5 + x)

A = 2(2x + 5)

A = 4x + 10

B = 2(l + w)

B = 2(x + 7 + x/2)

B = 2(3x/2 + 7)

B = 3x + 14

The perimeter of rectangle A is larger than that of rectangle B.

A > B

4x + 10 > 3x + 14

(4x + 10) - 3x > (3x + 14) - 3x

x + 10 > 14

(x + 10) - 10 > 14 - 10

x > 4

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A solid oblique pyramid has an equilateral triangle as a base with an edge length of 4 cm and an area of 12 cm2. What is the vol

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3 years ago

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Which expression is equivalent to [(3xy^-5)^3/(x^-2y^2)^-4]^-2?

Mademuasel [1]

Answer:- a.The given expression is equivalent to $\frac{x^{10}y^{14}}{729}$

Given expression:- $[\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}}]^{-2}$

$=[\frac{(3)^3x^3y^{-5\times3}}{x^{-2\times-4}y^{2\times-4}}]^{-2}.........(a^m)^n=a^{mn}$

$=[\frac{27x^3y^{-15}}{x^8y^{-8}}]^{-2}$

$=[27x^{3-8}y^{-15-(-8)}]^{-2}............\frac{a^m}{a^n}=a^{m-n}$

$=[27x^{-5}y^{-7}]^{-2}=(27)^{-2}(x^{-5})^{-2}(y^{-7})^{-2}.........(a^m)^n=a^{mn}$

$=\frac{1}{(27)^2}(x^{10}y^{14})=\frac{x^{10}y^{14}}{729}$

Thus a. is the right answer.

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3 years ago

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Pls help me and show working​

Pls help me and show working